Trigonometry / Unit 5: Inverse Trigonometric Functions
5.4 Compositions with Triangles
Notes
To evaluate something like $\displaystyle \cos(\arcsin\frac{3}{5})$, let $\displaystyle \theta = \arcsin\frac{3}{5}$ and draw a right triangle.
- $\displaystyle \sin\theta = \frac{3}{5}$ means opposite 3, hypotenuse 5, so the adjacent side is $\displaystyle \sqrt{25 - 9} = 4$.
- Then $\displaystyle \cos\theta = \frac{4}{5}$.
- Use the range of the inverse function to pick the quadrant and signs. Arcsine and arctangent give angles in Quadrants I and IV; arccosine gives Quadrants I and II.
Practice
What is $\displaystyle \cos\left(\arcsin\frac{3}{5}\right)$?
Opposite 3, hypotenuse 5, adjacent 4.
What is $\displaystyle \tan\left(\arccos\frac{5}{13}\right)$?
Adjacent 5, hypotenuse 13, opposite 12, so $\displaystyle \tan\theta = \frac{12}{5}$.
What is $\displaystyle \cos\left(\arctan\left(-\frac{4}{3}\right)\right)$?
The angle is in Quadrant IV with $x = 3$, $y = -4$, $r = 5$. Cosine is $\displaystyle \frac{x}{r} = \frac{3}{5}$.
Ready for more? Take the Unit 5 test.